boundary of a closed set is nowhere dense

Let A be closed. In general, the boundary of a set is closed. So it suffices to show that ∂⁡A has empty interior.

Let U⊂∂⁡A be open. Since ∂⁡A⊂A¯=A, this implies that U⊂A. Since int⁡(A) is the largest open subset of A, we must have U⊂int⁡(A). Therefore U⊂∂⁡A∩int⁡(A). But ∂⁡A∩int⁡(A)=(A¯-int⁡(A))∩int⁡(A)=∅, so U=∅.